Asked by sahmit
solve the equation
sin(3x) + (1/2) = 2cos^2(x)
for X to be between 0 and 360
sin(3x) + (1/2) = 2cos^2(x)
for X to be between 0 and 360
Answers
Answered by
MathMate
Use the relation
cos²(x)=1-sin²(x)
to convert the equation to a quadratic in sin(x).
Solve for the value of sin(x), say s1 and s2 and reject any impossible roots such as |s|>1.
For the remaining root(s), solve for
sin(θ)=s. There should always be two values of θ (0≤ θ<360°)for each valid value of s, where -1≤s≤+1.
If you need further help, post.
cos²(x)=1-sin²(x)
to convert the equation to a quadratic in sin(x).
Solve for the value of sin(x), say s1 and s2 and reject any impossible roots such as |s|>1.
For the remaining root(s), solve for
sin(θ)=s. There should always be two values of θ (0≤ θ<360°)for each valid value of s, where -1≤s≤+1.
If you need further help, post.
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